Inorganic and Applied Chemistry

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Inorganic and Applied Chemistry

Example 5- B:

pH calculation in weak acid solution

A solution consists of 1.0 M HF. HF has a Ka value of 7.2 · 10-4 M and we thereby have a weak acid

solution. We wish to calculate the pH-value of the solution.

The first step is to write the most important components in the solution which in this case is:

HF(aq) and H 2 O(l)

After this we write the aqueous equilibrium:

HF(aq) + H 2 O(l)  H 3 O+(aq) + F-(aq)

We return to the equilibrium expression which is written as:

  

HF

HO F

Ka M



    7 , 210

4 3

Analogously to prior examples we now pay attention to initial and equilibrium conditions. The initial and

end-concentrations are written and as in chapter 4 the parameter x gives the relations to the equilibrium

concentrations.

[HF] 0 = 1,0 M

[F-] 0 = 0 M

[H 3 O+] 0 = 10-7 M (from the autoprotolysis of water)

And end-concentrations by:

[HF] = (1,0 – x) M

[F-] = x M

[H 3 O+] = (10-7 + x) M

which gives the following equation:

x M

x

K M x x

a

2

7

(^42). 610

1. 
   


2. 210
   (^10 ) (
   ^
   Thereby [H 3 O+] = 2,6 · 10-2 M which is why the pH value in the solution becomes:
   pH logH 3 O log( 2. 610
   2 ) 1. 6
   Acids and bases